Prove the following trigonometric identities.
Question:

Prove the following trigonometric identities.

(i) $\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$

(ii) $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$

(iii) $\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\operatorname{cosec} \theta+\cot \theta$

(iv) $(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\sec \theta+\operatorname{cosec} \theta$

Solution:

(i) We have to prove the following identity-

$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$

Consider the LHS.

$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$

$=\left(\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}\right)\left(\frac{1+\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\right)$

$=\frac{(1+\cos \theta+\sin \theta)^{2}}{(1+\cos \theta)^{2}-\sin ^{2} \theta}$

$=\frac{2+2(\cos \theta+\sin \theta+\sin \theta \cos \theta)}{2 \cos ^{2} \theta+2 \cos \theta}$

$=\frac{2(1+\cos \theta)(1+\sin \theta)}{2 \cos \theta(1+\cos \theta)}$

$=\frac{1+\sin \theta}{\cos \theta}$

= RHS

Hence proved.

(ii) We have to prove the following identity-

$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$

Consider the LHS.

$\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}$

$=\left(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\right)\left(\frac{\sin \theta+\cos \theta+1}{\sin \theta+\cos \theta+1}\right)$

$=\frac{(\sin \theta+1)^{2}-\cos ^{2} \theta}{(\sin \theta+\cos \theta)^{2}-1}$

$=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2 \sin \theta \cos \theta}$

$=\frac{2 \sin \theta(1+\sin \theta)}{2 \sin \theta \cos \theta}$

$=\frac{1+\sin \theta}{\cos \theta}$

$=\left(\frac{1+\sin \theta}{\cos \theta}\right)\left(\frac{1-\sin \theta}{1-\sin \theta}\right)$

$=\frac{\cos \theta}{1-\sin \theta}$

$=\frac{1}{\sec \theta-\tan \theta}$ (Divide numerator and denominator by $\cos \theta$ )

RHS

Hence proved.

(iii) We have to prove the following identity-

$\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}=\operatorname{cosec} \theta+\cot \theta$

Consider the LHS.

$\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1}$

$=\frac{\cos \theta-\sin \theta+1}{\cos \theta+\sin \theta-1} \times \frac{\cos \theta+\sin \theta+1}{\cos \theta+\sin \theta+1}$

$=\frac{(\cos \theta+1)^{2}-(\sin \theta)^{2}}{(\cos \theta+\sin \theta)^{2}-(1)^{2}}$

$=\frac{\cos ^{2} \theta+1+2 \cos \theta-\sin ^{2} \theta}{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}$

$=\frac{\cos ^{2} \theta+1+2 \cos \theta-\left(1-\cos ^{2} \theta\right)}{1+2 \cos \theta \sin \theta-1}$

$=\frac{2 \cos ^{2} \theta+2 \cos \theta}{2 \cos \theta \sin \theta}$

$=\frac{2 \cos \theta(\cos \theta+1)}{2 \cos \theta \sin \theta}$

$=\frac{\cos \theta+1}{\sin \theta}$

$=\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}$

$=\cot \theta+\operatorname{cosec} \theta$

= RHS

Hence proved.

(iv)
Consider the LHS.

$(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$

$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$

$=(\sin \theta+\cos \theta)\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \times \cos \theta}\right)$

$=\frac{\sin \theta+\cos \theta}{\sin \theta \times \cos \theta} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$

$=\sec \theta+\operatorname{cosec} \theta$

= RHS

Hence proved.